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The probability that Mike can win a competition is 1/4, Rob

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Manager
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The probability that Mike can win a competition is 1/4, Rob [#permalink]

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New post 26 Oct 2006, 17:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The probability that Mike can win a competition is 1/4, Rob winning is 1/3and Ben winning is 1/6. What is the probability of Mike or Rob winning but not Ben



1/6
5/12
7/12
5/6
1
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New post 26 Oct 2006, 17:55
= (mike wins)and(rob wins)and(ben loses) OR (mike wins)and(rob loses)and(ben loses) OR (mike loses)and(rob wins)and(ben loses)

= (1/4)(1/3)(1-1/6) + (1/4)(1-1/3)(1-1/6) + (1-1/4)(1/3)(1-1/6)

= (1/4)(1/3)(5/6) + (1/4)(2/3)(5/6) + (3/4)(1/3)(5/6)

=5/6[ 1/12 + 2/12 + 3/12]

= 5/6[1/2]

= 5/12
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New post 26 Oct 2006, 20:21
1/4 + 1/3 - 1/6 = 5/12.


B is right.
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New post 26 Oct 2006, 21:02
Bad question, did not specified the possible winning condition. Who would think that a competition can be won by three people together.
  [#permalink] 26 Oct 2006, 21:02
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The probability that Mike can win a competition is 1/4, Rob

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